Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class Main {
    public static void main(String args[]) {
      Scanner sc = new Scanner(System.in);
      PrintWriter w = new PrintWriter(System.out);
      
      int n = sc.nextInt();
      double head_pro[] = new double[n+1];
      double tail_pro[] = new double[n+1];
      for(int i=1;i<=n;i++)
      {
          head_pro[i] = sc.nextDouble();
          tail_pro[i] = 1-head_pro[i]; 
      }
      double dp[][] = new double[n+1][n+1];
      dp[0][0] = 1;
      for(int i=1;i<=n;i++)
      {
          for(int j=0;j<=n;j++)
          {
              if(i<j)
                continue;
              if(j==0)
              {
                  dp[i][j]=dp[i-1][j]*tail_pro[i];
              }
              else
              {
                  dp[i][j]=(dp[i-1][j-1]*head_pro[i])+(dp[i-1][j]*tail_pro[i]);
              }
          }
      }
      double ans = 0;
      for(int i=n;i>n/2;i--)
      {
          ans+=dp[n][i];
      }
      w.println(ans);
      w.close();
    }
}